对矩阵A,求一可逆矩阵P,使P^TAP为对角矩阵（老师,用配方法解答）
问题描述：
对矩阵A,求一可逆矩阵P,使P^TAP为对角矩阵（老师,用配方法解答）
A=1 2 1
2 1 1
1 1 3
答
f = x1^2+x2^2+3x3^2+4x1x2+2x1x3+2x2x3
= (x1+2x2+x3)^2-3x2^2+2x3^2-2x2x3
= (x1+2x2+x3)^2-3(x2^2+(2/3)x2x3)+2x3^2
= (x1+2x2+x3)^2-3(x2+(1/3)x3)^2+(1/3)x3^2+2x3^2
= (x1+2x2+x3)^2-3(x2+(1/3)x3)^2+(7/3)x3^2
y1=x1+2x2+x3
y2=x2+(1/3)x3
y3=x3
C =
121
011/3
001
P=C^-1 即为所求.